Question

When ultraviolet rays incident on metal plate then photoelectric effect does not occur, it occurs by incidence of : -
A. Infrared rays
B. X-rays
C. Radio wave
D. Light wave

Answer 1

Hint:The photoelectric effect occurs when the energy carried by the incident photons is greater than the work function of the metal plate. The energy of the photon is proportional to the frequency of the photon. So, by comparing the frequencies of the given ultraviolet rays, we can determine for which ray the photoelectric effect may occur.

Formula used:
\[E = h\nu \]
where h is the Plank’s constant and E is the energy of the photon with frequency equals to \[\nu \].
\[c = \nu \lambda \]
where c is the speed of light, \[\nu \] is the frequency of the photon and \[\lambda \] is the wavelength of the light wave.

Complete step by step solution:
As the energy of the photon is directly proportional to the frequency of the wave so the energy of the visible light is more than that of infrared photons.
\[E \propto \nu \]
As it is given that the photoelectric effect doesn’t occur when the metal surface is illuminated with ultraviolet rays. This means that the frequency of the ultraviolet is less than the threshold frequency of the photoelectric metal. So, to show the photoelectric effect we need ray with higher frequency than the ultraviolet rays.

Light spectrum is the combination of electromagnetic waves of many different wavelengths of energy produced by the light source. As the frequency of the electromagnetic wave is the most characteristic property, the spectrum is characterized based on the range of the frequencies.

Electromagnetic rays	Range of frequency
\[\gamma - rays\]	\[{10^{20}} - {10^{24}}Hz\]
X-rays	\[{10^{17}} - {10^{20}}Hz\]
Ultraviolet rays	\[{10^{15}} - {10^{17}}Hz\]
Visible spectrum	\[{10^{14.5}} - {10^{15}}Hz\]
Infra-red	\[{10^{11}} - {10^{14.5}}Hz\]
Microwave	\[{10^9} - {10^{11}}Hz\]
FM radio wave	\[{10^6} - {10^9}Hz\]
AM radio wave	\[{10^5} - {10^6}Hz\]
Long radio wave	\[{10^0} - {10^5}Hz\]

The frequency of the X-rays is greater than the frequency of the ultraviolet rays. So, the energy of the photons of the X-rays is greater than that of ultraviolet rays. So, the photoelectric effect may occur when the photoelectric plate is incident with X-ray.

Therefore, the correct option is B.

Note: From the electromagnetic spectrum we can also compare the wavelengths of the rays to determine the rays which can cause the photoelectric effect. But the energy of the photon is inversely proportional to the wavelength.